UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Continuity at $x=0$ requires left-hand limit = right-hand limit = $f(0)$.Differentiability requires left-hand derivative = right-hand derivative.Explanation:Given $f(x) = x(\sqrt{x} - \sqrt{x+1})$.Check continuity at $x=0$:Left-hand limit: $\lim_{x\to 0^-} x(\sqrt{x}-\sqrt{x+1}) = 0$ (since $x \to 0$).Right-hand limit: $\lim_{x\to 0^+} x(\sqrt{x}-\sqrt{x+1}) = 0$.$f(0) = 0(\sqrt{0} - \sqrt{1}) = 0$.Thus LHL = RHL = $f(0)$, so $f$ is continuous at $x=0$.Check differentiability at $x=0$:Left-hand derivative: $\lim_{h\to 0^-} \frac{f(-h) - f(0)}{-h} = \lim_{h\to 0^-} \frac{-h(\sqrt{-h} - \sqrt{-h+1})}{-h} = \lim_{h\to 0^-} (\sqrt{-h} - \sqrt{1-h}) = 0 - 1 = -1$.Right-hand derivative: $\lim_{h\to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h\to 0^+} \frac{h(\sqrt{h} - \sqrt{h+1})}{h} = \lim_{h\to 0^+} (\sqrt{h} - \sqrt{h+1}) = 0 - 1 = -1$.LHD = RHD = $-1$, so $f$ is differentiable at $x=0$.Answer:Option B: differentiable at $x=0$.